Answer
$\frac{2}{3}(\sqrt{3x}-ln|(1+\sqrt{3x})|)+C$
Work Step by Step
$\int\frac{1}{1+\sqrt{3x}}dx$
let $\sqrt{3x}=(3x^{\frac{1}{2}})=u$
$du=\frac{1}{2}(3x)^{-\frac{1}{2}}(3)dx$
$du=\frac{3}{2}(3x)^{-\frac{1}{2}} dx=\frac{3}{2\sqrt{3x}}dx$
$du= \frac{3}{2u}dx$
$(2u)du=3dx$
$dx=\frac{(2u)du}{3}$
$\int\frac{1}{1+\sqrt{3x}}dx$
$=\int\frac{2u}{3(1+u}du$
$=\frac{2}{3}\int\frac{u}{1+u}du$
$=\frac{2}{3}\int\frac{1+u-1}{1+u}du$
$=\frac{2}{3}\int(1-\frac{1}{1+u})du$
$=\frac{2}{3}(u-ln|1+u|)+C$
$=\frac{2}{3}(\sqrt{3x}-ln|(1+\sqrt{3x})|)+C$
